$f\,^{\prime}(x)=9e^x$ and $f(8)=-8+9e^8$. $f(0) = $
Solution: Finding $f(x)$ We have $f'(x)=9e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (9e^x)\,dx \\\\ & = {9e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(8)=-8+9e^8$. Here's what we get when we plug in $8$ : $\begin{aligned}f(8)&={9e^{8}} {+ C} \end{aligned}$ We are given that this must equal $-8+9e^8$ : $-8+9e^8 = {9e^{8}} {+ C}$ Solving the equation gives us ${C=-8}$. Finding $f(0)$ Now, we have that $f(x)={9e^x} {-8}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=9e^0 - 8\\\\ &=1 \end{aligned}$ The answer $f(0) = 1$